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I know that a measurable space is a tuple $(X,\Sigma)$, where $X$ is a set and $\Sigma$ is a $\sigma$-algebra of (measurable) subsets of $X$. A measure space, in contrast is a triple $(X,\Sigma,\mu)$, where $X$ is a set, $\Sigma$ is a $\sigma$-algebra of (measurable) subsets of $X$, and $\mu$ is a measure.

The obvious main difference is that a measurable space does not require a specific measure. Does this mean that measurability as a concept is completely independent of any specific measure? Are $\sigma$-algebras always measurable by any measure, or are they considered measurable if at least one valid measure exists? If measurability is a concept independent of specific measures, why is there a correspondence between, for example, Borel $\sigma$-algebras and the Borel measure, or Lebesque $\sigma$-algebras and the Lebesgue measure?

Perhaps someone can clarify and point me to some resources that make this clear. Thank you!

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Given a pair $(X, \Sigma)$ consisting of a set $X$ and a $\sigma$-algebra $\Sigma$ of subsets of $X$, we frequently call the elements of $\Sigma$ measurable. However, this is just a name and does not tell you anything more (this is similar to calling the elements of a topology open subsets).

The point is, when you start with $(X, \Sigma)$, you declare $\Sigma$ to be the sets of subsets of $X$ which you want to measure, only the measure is missing and there are many measures which might be of interest. For example, the Lebesgue measure is only one of many measures which can be defined on the Lebesgue $\sigma$-algebra. Probability theory will give you many other important examples.

Then again, you can replace $\Sigma$ by another $\sigma$-algebra $\Sigma'$ to obtain a new measurable space $(X, \Sigma')$ in which sets which are measurable in $(X,\Sigma)$ might not be measurable any longer (and vice versa, of course).

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See the answers you have received allready. I would like to add that so-called outer measures induce $\sigma$-algebras. This in such a way that the the outer measure restricted to it is a measure on it. That might explain the use of terminology like "Lebesgue measure". We start with what we allready call "the Lebesgue outer measure", and the $\sigma$-algebra induced by it is called "Lebesgue $\sigma$-algebra". So the notion of measuring sets then precedes the notion of $\sigma$-algebra. See here and/or here for more information.

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    $\begingroup$ Great answer, because unlike the other answers it shows understanding for the student's (and OP's) mind that tries to approach new things from concepts it already knows and not from axioms and definitions. $\endgroup$ – Bananach Mar 7 '17 at 14:52
  • $\begingroup$ So can we for example define two measures $\mu$ and $\nu$ on the interval algebra such that $\sigma$-algebras generated by $\mu$ and $\nu$ do not coincide? $\endgroup$ – Daron Mar 7 '17 at 16:11
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    $\begingroup$ @Daron Yes. If $\lambda^*$ denotes the Lebesgue outer measure on $\mathbb R$ then (under AC) the generated $\sigma$-algebra can be shown to be a proper subset of $\wp(\mathbb R)$. If $\mu^*$ is defined by $\mu^*(I)=0$ for every interval $I$ then the generated $\sigma$-algebra coincides with $\wp(\mathbb R)$. $\endgroup$ – drhab Mar 7 '17 at 16:35
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Yes, a measurable space $(X, \Sigma, \mu)$ means that $(X, \Sigma)$ must be a $\sigma$-algebra. The function $\mu$ is required to be a function defined on the entire $\Sigma$.

So the measurability in a measurable space depends on the corresponding $\sigma$-algebra. That is $E$ is measurable in $(X, \Sigma, \mu_1)$ iff it is measurable in $(X, \Sigma)$ which it is iff it's measurable in $(X, \Sigma, \mu_2)$.

Of course measurability is not more universal than that. We can have $(X, \Sigma_1)$ and $(X, \Sigma_2)$ with $\Sigma_1 \ne \Sigma_2$ whenever $|X|\ge 2$.

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